Ncert Exercises & Solutions

100 g of water is supercooled to —10°C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the témnperature of the resultant mixture and how much mass would freeze? $[S_{w} = 1 c a l / g / ° C a n d L_{F u s i o n}^{w} = 80 c a l / g]$

Hint: Apply the concept of calorimetry.

Step 1: Find the heat required to increase the temperature of the water.

Given, the mass of water, (m)= 100
Change in temperature,

∆ T = 0 - (- 10) = 10 ° C

Specific heat of water,

(s_{w}) = 1 c a l / g / ° C

Latent heat of fusion of water,

L_{f u s i o n_{w}} = 80 c a l / g

The heat required to bring water in super cooling from

- 10 ° C t o 0 ° C,

Q = m s_{w} ∆ T = 100 \times 1 \times 10 = 1000 c a l

Step 2: Find the amount of ice melted.

Let m gram of ice be melted.

∴ Q = m L

m = \frac{Q}{L} = \frac{1000}{80} = 12.5 g

As the small mass of ice is melted, therefore the temperature of the mixture will remain 0°C.

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